12 volt Electrical Battery Problems. Where is all the power going?

[roaringgirl]

It sounds like you have a shunt that everything runs through - which means you can trust the monitor to tell you accurately the net flow of current. I'm sure you know this - the useable capacity of your batteries is about 50% of the nominal, so your 110Ah has 55 useable Ah in it. If your 2 fridges are constantly drawing 8A each then your batt monitor will read 16A and your 110Ah battery will be drained in 3.4 hours (55/16 = 3.4).

Did you say that you have 2 110Ah batteries, and they're drained in 6h? - That sounds about right, based on the above calculations assuming no input. If you know each fridge draws 8A, then your solar is giving you 16A-whatever your battery monitor reads.

So here's the equation for what you can reasonably expect:

Number of hours to flatten batteries = ([Nominal capacity]/2)/[shunt current drain reading]
What does your shunt monitor read?
Is your nominal capacity 220Ah?

 

[VicS]

Which Mobicool 40 do you have?

presumably the FR 40 if it is a compressor type

If so it has an adjustable temperature setting and the OP could make a worthwhile saving in power consumption by settin it at a sensible value instead running it minus 4 C

 

[pvb]

It sounds like you have a shunt that everything runs through - which means you can trust the monitor to tell you accurately the net flow of current. I'm sure you know this - the useable capacity of your batteries is about 50% of the nominal, so your 110Ah has 55 useable Ah in it. If your 2 fridges are constantly drawing 8A each then your batt monitor will read 16A and your 110Ah battery will be drained in 3.4 hours (55/16 = 3.4).

Did you say that you have 2 110Ah batteries, and they're drained in 6h? - That sounds about right, based on the above calculations assuming no input. If you know each fridge draws 8A, then your solar is giving you 16A-whatever your battery monitor reads.

So here's the equation for what you can reasonably expect:

Number of hours to flatten batteries = ([Nominal capacity]/2)/[shunt current drain reading]
What does your shunt monitor read?
Is your nominal capacity 220Ah?

I'm afraid that a lot of that isn't true.

A 110Ah battery discharged at 16A will be down to 50% after about 1.5 hours, not 3.4 hours. This is because of the Peukert effect, that the effective capacity of a battery reduces as the discharge current increases. You can find online calculators which will help you to understand this - try this one for starters http://www.csgnetwork.com/batterylifecalc.html The Peukert number for an average flooded battery is about 1.3; the Peukert number for an AGM battery is around 1.1.

 

[vyv_cox]

For reference, my experience.

My fridge runs on a Danfoss BD35 compressor modified for seawater cooling. It draws just over 4 amps, including about 0.5 amps for the seawater pump, at the start of its cycle, reducing to just over 3 amps at the end. When I last checked it was doing about 10 minutes running, 20 minutes off, in water temperature of about 26C. I have 125 watts of solar power and 3 x 110 deep cycle batteries. In June/July we are totally self sufficient with this arrangement but need to run the engine briefly for charging in September/October.

When I only had a domestic battery bank of 2 x 110 Ah the voltage was down to just over 12 volts first thing in the morning, whereas with 3 x 110 Ah it is usually 12.4 volts in mid-summer.

This suggests that running two fridges with only two domestic batteries is pushing things a little too far, although it sounds as if there is an additional problem, perhaps to do with the initial charge level. I cannot comment on the charge monitor as I have never used one.

 

[halcyon]

.

When I only had a domestic battery bank of 2 x 110 Ah the voltage was down to just over 12 volts first thing in the morning, whereas with 3 x 110 Ah it is usually 12.4 volts in mid-summer.

This suggests that running two fridges with only two domestic batteries is pushing things a little too far, although it sounds as if there is an additional problem, perhaps to do with the initial charge level. I cannot comment on the charge monitor as I have never used one.

Without facts, you only have supposition, does the fridge have a thermostat? what current during running on time ? what time does it run for ? drawing 8 amps continuous is totally different to 8 amps for 10 minutes every hour.

Brian

 

[vyv_cox]

Without facts, you only have supposition, does the fridge have a thermostat? what current during running on time ? what time does it run for ? drawing 8 amps continuous is totally different to 8 amps for 10 minutes every hour.

Brian

Not sure why you have posted this. The facts are in the paragraph above the one you have quoted. What other facts do you want? I do not know how the OP's fridges are cooled but assume air, in which case they will run longer than mine.

 

[halcyon]

Not sure why you have posted this. The facts are in the paragraph above the one you have quoted. What other facts do you want? I do not know how the OP's fridges are cooled but assume air, in which case they will run longer than mine.

I do not know how the OP's fridges are cooled but assume air, in which case they will run longer than mine

My point.

To be honest I do not know why I bother either, got two switch panels to get designed.

Brian

 

[Daverw]

Just to add that if your running your mobicool at -4 you are spending loads of power frying to freeze your drinks rather than just keeping them cool, set if for the temp you want the drinks at and don’t use it as a blast chiller

 

[roaringgirl]

I'm afraid that a lot of that isn't true.



A 110Ah battery discharged at 16A will be down to 50% after about 1.5 hours, not 3.4 hours. This is because of the Peukert effect, that the effective capacity of a battery reduces as the discharge current increases. You can find online calculators which will help you to understand this - try this one for starters http://www.csgnetwork.com/batterylifecalc.html The Peukert number for an average flooded battery is about 1.3; the Peukert number for an AGM battery is around 1.1.

Read it again, i said that 2x110Ah will be drained by 16A in ~3.4hrs~. !! Correction !! about 6 hrs.

 

[pvb]

Read it again, i said that 2x110Ah will be drained by 16A in 3.4hrs.

I don't quite believe this! You try reading again what you wrote, which was "If your 2 fridges are constantly drawing 8A each then your batt monitor will read 16A and your 110Ah battery will be drained in 3.4 hours (55/16 = 3.4)." Are you capable of understanding what you wrote?

 

[roaringgirl]

I don't quite believe this! You try reading again what you wrote, which was "If your 2 fridges are constantly drawing 8A each then your batt monitor will read 16A and your 110Ah battery will be drained in 3.4 hours (55/16 = 3.4)." Are you capable of understanding what you wrote?

1) Approximate real world capacity = Nominal capacity/2 = 55Ah
2) Current drain from 2 8A fridges = 16A
3) Hours until drained = [Approx real world capacity]/[Current drain] = 55Ah/16A = 3.4Hrs.

Which statement don't you like? Whichever it is, see if you can disagree with it without making it personal...

 

[Polly's Kettle]

Would you be willing to operate without your fridges for a day?

Why not simulate the same event by removing yourself from shore power, but switch off everything drawing power. If the batteries drop off at all, then you've confirmed they are dead or that the solar panels are delivering nothing. Then turn on everything else that you use except the fridges, and see if you get anything like your usual plummet.

Then turn on each of the fridges separately, and see what happens. Then do both. You need to test the various scenarios to make sure you know what is doing the damage. It would be a shame to put in new batteries and kill them because you had a faulty fridge. or faulty something else not yet identified.

 

[pvb]

1) Approximate real world capacity = Nominal capacity/2 = 55Ah
2) Current drain from 2 8A fridges = 16A
3) Hours until drained = [Approx real world capacity]/[Current drain] = 55Ah/16A = 3.4Hrs.

Which statement don't you like? Whichever it is, see if you can disagree with it without making it personal...

Statement 3 is completely incorrect. I explained why in post 23, and said that the battery would be drained in about 1.5 hours, not 3.4 hours.

As you seem not to understand battery theory, it would be better if you didn't post incorrect advice.

 

[noelex]

1) Approximate real world capacity = Nominal capacity/2 = 55Ah
2) Current drain from 2 8A fridges = 16A
3) Hours until drained = [Approx real world capacity]/[Current drain] = 55Ah/16A = 3.4Hrs.

Which statement don't you like? Whichever it is, see if you can disagree with it without making it personal...

Roaringgirl, your calculations above do not take into account Peukets equation. This is often incorrectly ignored, but does need to be taken into account, particularly when the discharge is high relative to battery size.

Pvb is trying to point this out in his post.

Unfortunately, we cannot do an exact calculation taking into account Peukets equation unless we know the specifics of the battery, which needs to be supplied by the battery manufacturer.

A rough “back of the envelope” calculation suggests the duration for a battery rated at 110Ahr @C20 would be about 2 -2.5 hours. Pvb’s estimation of 1.5 hours is I think incorrect. This sounds way too low a capacity that is estimated from a discharge rate than puts the capacity above the C5 rate. So I think something has gone wrong with the maths in making this estimation.

So therefore I think you are both wrong and the correct answer is somewhere in the middle of both of your estimations .

 

[pvb]

A rough “back of the envelope” calculation suggests the duration for a battery rated at 110Ahr @C20 would be about 2 -2.5 hours. Pvb’s estimation of 1.5 hours is I think incorrect. This sounds way too low a capacity that is estimated from a discharge rate than puts the capacity above the C5 rate. So I think something has gone wrong with the maths in making this estimation.

So therefore I think you are both wrong and the correct answer is somewhere in the middle of both of your estimations .

The 1.5 hours is the time to reach 50% discharged (the notional level which it's safe to go to on a regular basis). This is for an average flooded battery with a Peukert number of 1.3. If you think this is incorrect you should take it up with the people who have the online calculator I linked to in post 23, but you'll be wasting your time because all the online calculators will come up with the same sort of figure. A different sort of battery will give a different 50% discharge time; for example an AGM battery with a Peukert number of 1.1 would last 2.6 hours.

 

[roaringgirl]

Statement 3 is completely incorrect. I explained why in post 23, and said that the battery would be drained in about 1.5 hours, not 3.4 hours.

As you seem not to understand battery theory, it would be better if you didn't post incorrect advice.

My approximation comes to the same conclusion as yours - the batteries are performing as expected by being emptied quite quickly. You seem a bit upset about something, I hope you manage to get better



The next question then, is why are both fridges drawing 8A each all the time - are the thermostats broken?

 

[pvb]

My approximation comes to the same conclusion as yours - the batteries are performing as expected by being emptied quite quickly.

If you think 3.4 hours is the same as 1.5 hours, you're taking "approximation" to a new level!

The next question then, is why are both fridges drawing 8A each all the time - are the thermostats broken?

They're not drawing 8A each, the OP said "..together they take 8 amps approximately". It always helps to read the OP.

 

[noelex]

If you think this is incorrect you should take it up with the people who have the online calculator I linked to in post 23, but you'll be wasting your time because all the online calculators will come up with the same sort of figure.

Sorry, I did not see your link. Unfortunately, the calculation in the link is incorrect (or at least it assumes the wrong starting parameters). It assumes a battery is rated with a 1 amp discharge. All marine batteries are rated at C20 (or very occasionally C100).

For example: If you put into the table a 110 Ahr battery with a 5.5A discharge, the table gives an answer of 66AHrs when of course the correct answer is 110Ahrs because this is how the battery is rated.

 

[pvb]

Sorry, I did not see your link. Unfortunately, the calculation in the link is incorrect (or at least it assumes the wrong starting parameters). It assumes a battery is rated with a 1 amp discharge. All marine batteries are rated at C20 (or very occasionally C100).

For example: If you put into the table a 110 Ahr battery with a 5.5A discharge, the table gives an answer of 66AHrs when of course the correct answer is 110Ahrs because this is how the battery is rated.

OK I just tried another calculator, which came up with 2.5 hours to 50%, same as your guesstimate but still way less than the 3.4 hours believed by roaringgirl. The key point for people to understand is that the capacity of a battery reduces quite significantly when discharged at higher currents than the C20 rate; you can't just divide capacity by current and get an accurate run time.

 

[temptress]

Forget about the % shown. What is the voltage measured at the batteries when you are seeing the 8amp draw. Also what else are you running? STERRO lights what is the actual amp draw and what voltage do you see at the batteries?

Start there...